Physics 1151
Exam #3
November 15, 2002

Possibly useful equations:

My powers of intuition and guessing are non-existent. Please make your answers clear and easy to follow. Unlabeled numbers make it very difficult to guess or figure out what you are doing. Tell me the details! Label and explain what you are doing.

| Return to 1151 Calendar |

1. (9.29) A solid, uniform cylinder of 12 cm radius with mass of 5.0 kg is free to rotate about its symmetry axis. A cord is wound around the drum and a 1.2 kg mass is attached to the end of the cord. Find the acceleration of the hanging mass, the angular acceleration of the cylinder, and the tension in the cord.

For the 1.2 kg mass, take down as positive and apply Newton's Second Law,

Fnet = m a
Fnet = 12 N - T = (1.2 kg) a
12 N - T = (1.2 kg) a

For the rotating cylinder, apply the rotational form of Newton's Second Law,

net = I

net = r F sin
net = (0.12 m) (T) (1)

since = 90°

For a solid cylinder,

I = (1/2) M r2
I = (0.5) (5.0 kg) (0.12 m)2
I = 0.036 kg m2


net = (0.12 m) T = (0.036 kg m2)

At the moment, we have three unknowns, T, a, and , in two equations. The angular acceleration of the cylinder and the linear acceleration of the cord or of the hanging mass are connected by

a = r
a = (0.12 m)
= a / 0.12 m

(0.12 m) T = (0.036 kg m2)

(0.12 m) T = (0.036 kg m2) [ a / 0.12 m]
T = (0.036 kg m2) a/[ (0.12 m)2(0.12 m) ]

T = (0.036 kg m2) a/(0.12 m)2
T = (2.5 kg) a

Now we have two equations in two unknowns

12 N - T = (1.2 kg) a
12 N - (2.5 kg) a = (1.2 kg) a
12 N - (2.5 kg) a = (1.2 kg) a
12 N = (3.7 kg) a
a = 12 N / 3.2 kg
a = 3.24 m/s2

T = 8.1 N

2. (10.56) What percentage of a copper cube floats above the surface if it floats in mercury?

(Hg) = 13,600 kg/m3

(Cu) = 8,920 kg/m3

3. (11.6) What must be the gauge pressure at a fire hydrant if water from a fire hose is to reach a height of 12 m?

We can immediately apply Bernoulii’s Equation,

1/2 v2 + g h + p = constant


1/2 v12 + g h1 + p1 = 1/2 v22 + g h2 + p2

In our case, the hydrant is position 1 and the top of the stream of water is position 2,

v1 = 0

h1 = 0

p1 = p1g = ?


v2 = 0

h2 = 12 m

p2 = p2g = 0

(The guage pressure at the top of the stream is zero).

1/2 (0)2 + g (0) + p1g = 1/2 (0)2 + g (12 m) + (0)
p1g = g (12 m)
p1g = (1000 kg/m3) (9.8 m/s2 ) (12 m)
p1g = 117,600 N/m3 = 117,600 Pa = 117.6 kPa
p1g = 117.6 kPa
p1g = 117.6 kPa [ 1 atm/101.3 kPa ] = 1.16 atm

Remember, the subscript "g" simply means "gauge pressure".

4. (12.57) A 0.050-kg ice cube, initially at – 5.0°C, is placed in 0.30 kg of water at 25°C.

What is the final temperature of the water?

Or, if melting is not complete, how much ice remains unmelted at thermal equilibrium?

Assume no heat is lost to the environment.

First, assume all the ice melts, so Tf > 0°C
Q1 + Q2 = 0
Q1 = heat lost by the water
Q1 = cwater mwaterTwater = (4186 J/kg C°)(0.30 kg)(Tf – 25°C)
Q1 = (4186 J/kg C°)(0.30 kg)(Tf – 25°C) = 1,256 (J/) Tf – 31,395 J

Q2 = heat gained by ice
Q2 = heat gained in raising ice to 0°C +

+ heat gained in melting ice +

+ heat gained in raising the melted ice (now water) to temperature Tf

Q2 = cice mice T1 + Lf,ice mice + cwater mice DT2
Q2 = [2090 J/kg C°][0.05 kg][0°C – (– 5°C)] +
+ [3.33 x 105 J/kg][0.05 kg] +
+ [4186 J/kg C°][0.05 kg][Tf – 0°C]
Be very careful with these temperature changes !
Q2 = 522.5 J + 16,650 J + 209.3 (J/) Tf – 0
Q2 = 17,172.5 J + 209.3 (J/) Tf

Now we are ready to go back to
Q1 + Q2 = 0
1,256 (J/) Tf – 31,395 J + 17,172.5 J + 209.3 (J/) Tf = 0
1,256 (J/) Tf + 209.3 (J/) Tf = 31,395 J – 17,172.5 J
1,465 (J/) Tf = 14,222 J
Tf = [14,222/1,465]°C
Tf = 9.7°C

5. Concept Questions:

Consider a twirlling ice skater (or a dizzy Physics professor) who spends faster as she (or he) pulls her (or his) hands in close. Spinning increases because angular momentum is conserved. However, the rotational kinetic energy increases. This increase in kinetic energy means that work was done. Where or how is work done in this situation to increase the kinetic energy?

An inward force is required to pull the weights inward. That force is supplied by the arms of the rotating skater (or professor) and acts through a distance so that provides the work done to increase the kinetic energy.

ii) Consider an ocean liner that leaves l’Harve, France, crosses the Atlantic Ocean, enters the St Lawerence River, and docks at Montreal, Quebec. The Atlantic Ocean is sea water and the St Lawerence is fresh water. What happens to the draft of the ocean liner as it goes from the salt water of the Atlantic to the fresh water of the St Lawerence? Why? The “draft” of a ship is the depth of the ship below the water.

The bouyant force must equal the weight of the vessel. The bouyant force is provided by the weight of the water displaced. Since sea water has a greater density, less sea water must be displaced. That means the draft in fresh water will be greater (deeper) than the draft in sea water.

iii) In class, I demonstrated a ping-pong ball in a funnel. As I blew air through the funnel, the ping-pong ball was held in place. Why? Why was it not blown out of the funnel?

The Bernoulli effect is that moving air exerts less pressure than still air. The moving air between the funnel and the ping-pong ball exerts less pressure than the still air on the other side of the ping-pong ball. The greater pressure on the other side of the ping-pong ball means there is greater force there, pushing the ball into the funnel.

iv) Name a thermometric property and describe how it can be used in a thermometer?

A thermometric property is a property that changes with the temperature.

As a liquid is heated, its volume expands. This can be used to make a common liquid-in-glass thermometer.

As a metal is heated, its length increases. This can be used to make a bimetallic strip thermometer.

As a resistor is heated, its resistance changes. This can be used to make an electronic thermometer.

As a gas is heated, its volume expands. This can be used to make a constant volume gas thermometer. We used such a thermometer to come up with the absolute temperature scale, now more usually known as the Kelvin temperature scale.

| Return to 1151 Calendar |